What guarantees the existence of unitary operators implementing Lorentz Transformations? This should be a very basic question. In introductory QFT books, often one of the first things we see is the following claim: for every Lorentz transformation $\Lambda$, we can associate an unitary operator $U(\Lambda)$ such that:
$$ U(\Lambda)^{-1} \varphi(x) U(\Lambda)= \varphi(\Lambda^{-1}x)$$
And we also demand it to be a homomorphism, $U(A)U(B)=U(AB)$.
Where of course, $\varphi$ is an operator-valued quantum field. 
I want to know what guarantees the existence of mappings $U$ which satisfy these conditions.
It seems like it ought be possible to choose an operator field $\varphi$ such that no such set of operators $U(\Lambda)$ exist. Is the existence of $U$ given by some requirement on quantum fields, or am I missing something else?
 A: Your last sentence answers your question.
We observe Lorentz symmetry in the laws of nature. Therefore, we demand that the building blocks of our theory transform in definite representations of the Lorentz (or rather Poincaré) group.
Would you allow fields that are not representations of the Lorentz group, it would become extremely hard to construct a theory that looks Lorentz invariant.
A: I claim that

In any relativistic quantum theory, Lorentz transformations of states must be realized as a projective, unitary representation of the Lorentz group acting on the Hilbert space of the theory.

Here's the logic:


*

*In any relativistic theory, the observations of spacetime events of inertial observers are related by Lorentz transformations.

*We ask ourselves how the observations of quantum states, namely elements of some Hilbert space $\mathcal H$ that models a certain quantum system, of different inertial observers are related.  In mathematical terms, for each Lorentz transformation $\Lambda$, we would like to associate a function $f_\Lambda:\mathcal H\to\mathcal H$ such that if $|\psi\rangle$ is the state measured by one inertial observer, then $f_\Lambda |\psi\rangle$ is the state measured by the inertial observer whose spacetime observations are related to the first by the Lorentz transformation $\Lambda$.

*We notice that whatever $f_\Lambda$ is, it should preserve quantum-mechanical transition probabilities.  In other words, for each $\Lambda$, $f_\Lambda$ should be a symmetry in the general quantum mechanical sense defined by the preservation of transition probabilities.

*We recall that Wigner's Theorem guarantees that any such symmetry can be represented by a unitary or anti-unitary operator up to phase.

*We argue no $f_\Lambda$ can be anti-unitary (I actually have forgotten the conventional argument for this, perhaps you can try to fill in this detail before I do).  So we use the notation $f_\Lambda = U(\Lambda)$ instead to emphasis this.

*We consider three inertial observers $A$, $B$, and $C$.  We let $\Lambda_{ij}$ be the Lorentz transformation that connects the spacetime measurements of observer $j$ to those of $i$, so, for example, $\Lambda_{AB}$ is the Lorentz transformation connecting the spacetime observation of observer $B$ to those of $A$.

*We note that the state measurements of observers $A$ and $C$ are related by $U(\Lambda_{AC})$.  Moreover, we expect that if we transform the state measurements of observer $A$ to those of $B$ with $U(\Lambda_{AB})$, and then transform those measurements to those of $C$ with $U(\Lambda_{BC})$, then we should get the same answer up to phase (aka a physically equivalent state), namely
\begin{align}
  U(\Lambda_{AC}) = c(\Lambda_{BC}, \Lambda_{AB})U(\Lambda_{BC})U(\Lambda_{AB}).
\end{align}
where the "up to phase" part comes from the fact that states that differ by a phase are physically equivalent in quantum mechanics.  But now we note that $\Lambda_{AC} = \Lambda_{BC}\Lambda_{AB}$, so we get the homomorphism property up to phase
\begin{align}
  U(\Lambda_{BC}\Lambda_{AB}) = c(\Lambda_{BC}, \Lambda_{AB})U(\Lambda_{BC})U(\Lambda_{AB})
\end{align}
and we're therefore done.

*Related post (to better understand the "projective" part): Idea of Covering Group 
